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A wire with a weight per unit length of 0.075 N/m is suspended directly above a second wire. The top wire carries a current of 29.2 A and the bottom wire carries a current of 59.1 A. Find the distance of separation between the wires so that the top wire will be held in place by magnetic repulsion.

Answer :

Answer:

Explanation:

force between two current carrying conductor per unit length

= (μ₀ / 4π) x (2 i₁i₂ / R)  , i₁ and i₂ be current in them , R is distance between two ,

= 10⁻⁷ x 2 x 29.2 x 59.1 / R .

This force balances weight of wire per unit length

= 10⁻⁷ x 2 x 29.2 x 59.1 / R . = .075

= 10⁻⁷ x 2 x 29.2 x 59.1 / .075 = R

R = 4.6 x 10⁻³ m

4.6 mm .

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